Variance of Chi Square Distribution as the Sum of Unit Normal Random Variables Okay, so I am interested if there is a way to derive the variance for a Chi-Square distribution using the property that it is the sum of independent unit normal distributions squared. 
For example, if $X$ is a Chi-Square random variable with $n$ degrees of freedom, it has the distribution: $\displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ where $Z$ is Normal$(0,1)$
I know that $Var(X)= E(X^{2}) -[E(X)]^{2}$
To start finding $E(X)$ I begin with the fact that each $Z$ has $E(Z)=0$ and $Var(Z)=1$. This implies that $E(Z^2)=1$ since $Var(Z)=E(Z^{2})-[E(Z)]^2$
Since $X = \displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ then $E(X)=\displaystyle\operatorname{}\left(\sum_{i=1}^n 1\right)=n$
I am lost on what the next step would be. I have this start, but don't where to go next. Any thoughts on how to find $E(X^{2})$?
$\begin{align}
X^2 =& (\sum_{i=1}^nZ_i^2)^2\\
=& \sum_{i=1}^nZ_i^4+\sum_{i \neq j}^nZ_i^2Z_j^2
\end{align}$ 
I can't see any way this sum not getting nasty. Note: I can solve this using integration of the PDF for the Chi-Square distribution, but I was wondering if there is any way to do it using the property that Chi-Squared is sum of Squared Normal.
 A: Since $X_1^2,\ldots,X_n^2$ are independent, you have
$$
\operatorname{var}(X_1^2+\cdots+X_n^2) = n \operatorname{var}(X_1^2).
$$
And this is $\operatorname{var}(X_1^2) = \operatorname{E}((X_1^2)^2) - (\operatorname{E}(X_1^2))^2$. If you know that $\operatorname{E}(X_1^2) =1,$ then it remains only to find $\operatorname{E}(X_1^4).$
\begin{align}
\operatorname{E}(X_1^4) & = \int_{-\infty}^\infty x^4 \varphi(x)\, dx \text{ where $\varphi$ is the standard normal density} \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2} \,dx \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty x^4 e^{-x^2/2} \,dx\text{ by symmetry} \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty x^3 e^{-x^2/2} \Big( x\,dx\Big) \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du = \frac 4 {\sqrt{\pi}} \int_0^\infty u^{3/2} e^{-u} \,du \\[10pt]
& = \frac 4 {\sqrt\pi} \Gamma\left(\frac 5 2 \right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \Gamma\left( \frac 3 2 \right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \Gamma\left( \frac 1 2 \right) \\[10pt]
& = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \sqrt \pi \\[10pt]
& = 3. 
\end{align}
